How do you solve $\frac{1}{4} x^{2} = 3 x - \frac{25}{2}$ $\frac { 1} { 4} x ^ { 2} = 3x - \frac { 25} { 2}$ ?

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Answer 1 · 2018-06-11T15:15:43

no solutions $ϕ$ $phi$

$\frac{1}{4} x^{2} = 3 x - \frac{25}{2}$ $\frac { 1} { 4} x ^ { 2} = 3x - \frac { 25} { 2}$

multiply both sides by 4 to remove the fractions:

$4 [\frac{1}{4} x^{2} = 3 x - \frac{25}{2}]$ $4[\frac { 1} { 4} x ^ { 2} = 3x - \frac { 25} { 2}]$

$x^{2} = 12 x - 50$ $x^2=12x-50$

$x^{2} - 12 x + 50 = 0$ $x^2-12x+50=0$

If we check the discriminant of the form $a x^{2} + b x + c$ $ax^2+bx+c$ :

$b^{2} - 4 a c$ $b^2-4ac$

${(- 12)}^{2} - 4 \cdot 1 \cdot 50 = - 56$ $(-12)^2-4*1*50=-56$

Since the discriminant is negative there are no real solutions to this polynomial as can be seen from its graph:

graph{x^2-12x+50 [-35.17, 44.83, -3.52, 36.48]}

How do you solve 14x2=3x−252\frac { 1} { 4} x ^ { 2} = 3x - \frac { 25} { 2}?